Given a point P in 2-D plane and equation of mirror, the task is to find the image of that point Q formed due to the mirror.
Equation of mirror is in form ax + by + c = 0
Examples:
Input : P = (1, 0), a = -1, b = 1, c = 0
Output : Q = (0, 1)Input : P = (3, 3), a = 0, b = 1, c = -2
Output : Q = (3, 1)
Solution :
Let coordinate of P(given point) be (x1, y1) Let coordinate of Q(image point) be (x2, y2) Let coordinate of R(point on mirror) be (x3, y3)
Since object and image are equidistant from mirror, R must be middle point of P and Q
Since equation of mirror is given to be: ax + by + c = 0. Equation of line passing through P and Q is perpendicular to mirror. Therefore equation of line passing through P and Q becomes ay – bx + d = 0, Also P passes through line passing through P and Q, so we put coordinate of P in above equation,
a*y1 – b*x1 + d = 0
d = b*x1 – a*y1
Also R is intersection of mirror and line passing through P and Q. So we find solution of
ax + by + c = 0
ay -bx + d = 0
Since a, b, c, d all are known we can find x and y here. Since coordinates of R is known now ie.. x3, y3 are known now.
Since R is middle point of PQ,
(x3, y3) = ((x1+x2)/2, (y1+y2)/2)
Since x1, y1, x3, y3 are known, we get below equation where (x, y) are coordinates of Q(image point)
We use above formula to find mirror image of point P(x1, y1) with respect to mirror of equation ax + by + c
C++
// C++ code to find mirror image#include <iostream>using namespace std;// C++ function which finds coordinates// of mirror image.// This function return a pair of doublepair<double, double> mirrorImage( double a, double b, double c, double x1, double y1){ double temp = -2 * (a * x1 + b * y1 + c) / (a * a + b * b); double x = temp * a + x1; double y = temp * b + y1; return make_pair(x, y);}// Driver code to test above functionint main(){ double a = -1.0; double b = 1.0; double c = 0.0; double x1 = 1.0; double y1 = 0.0; pair<double, double> image = mirrorImage(a, b, c, x1, y1); cout << "Image of point (" << x1 << ", " << y1 << ") "; cout << "by mirror (" << a << ")x + (" << b << ")y + (" << c << ") = 0, is :"; cout << "(" << image.first << ", " << image.second << ")" << endl; return 0;} |
Java
// Java code to find mirror imageclass GFG{static class pair{ double first, second; public pair(double first, double second) { this.first = first; this.second = second; } } // function which finds coordinates// of mirror image.// This function return a pair of doublestatic pair mirrorImage(double a, double b, double c, double x1, double y1){ double temp = -2 * (a * x1 + b * y1 + c) / (a * a + b * b); double x = temp * a + x1; double y = temp * b + y1; return new pair(x, y);}// Driver codepublic static void main(String []args){ double a = -1.0; double b = 1.0; double c = 0.0; double x1 = 1.0; double y1 = 0.0; pair image = mirrorImage(a, b, c, x1, y1); System.out.print("Image of point (" + x1 + ", " + y1 + ") "); System.out.print("by mirror (" + a + ")x + (" + b + ")y + (" + c + ") = 0, is :"); System.out.println("(" + image.first + ", " + image.second + ")");}}// This code is contributed by 29AjayKumar |
Python 3
# Python 3 code to find mirror image # Python function which finds coordinates # of mirror image. # This function return a pair of double def mirrorImage( a, b, c, x1, y1): temp = -2 * (a * x1 + b * y1 + c) /(a * a + b * b) x = temp * a + x1 y = temp * b + y1 return (x, y)# Driver code to test above function a = -1.0b = 1.0c = 0.0x1 = 1.0y1 = 0.0x, y = mirrorImage(a, b, c, x1, y1); print("Image of point (" + str (x1) + ", " + str( y1) + ") ")print("by mirror (" + str (a) + ")x + (" + str( b) + ")y + (" +str(c) + ") = 0, is :") print( "(" + str(x) + ", " + str(y) + ")" ) # This code is contributed by ApurvaRaj |
C#
// C# code to find mirror imageusing System;class GFG{class pair{ public double first, second; public pair(double first, double second) { this.first = first; this.second = second; } } // function which finds coordinates// of mirror image.// This function return a pair of doublestatic pair mirrorImage(double a, double b, double c, double x1, double y1){ double temp = -2 * (a * x1 + b * y1 + c) / (a * a + b * b); double x = temp * a + x1; double y = temp * b + y1; return new pair(x, y);}// Driver codepublic static void Main(String []args){ double a = -1.0; double b = 1.0; double c = 0.0; double x1 = 1.0; double y1 = 0.0; pair image = mirrorImage(a, b, c, x1, y1); Console.Write("Image of point (" + x1 + ", " + y1 + ") "); Console.Write("by mirror (" + a + ")x + (" + b + ")y + (" + c + ") = 0, is :"); Console.WriteLine("(" + image.first + ", " + image.second + ")");}}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript code to find mirror image // JavaScript function which finds coordinates // of mirror image. // This function return a pair of double function mirrorImage(a, b, c, x1, y1) { var temp = (-2 * (a * x1 + b * y1 + c)) / (a * a + b * b); var x = temp * a + x1; var y = temp * b + y1; return [x, y]; } // Driver code to test above function var a = -1.0; var b = 1.0; var c = 0.0; var x1 = 1.0; var y1 = 0.0; var [x, y] = mirrorImage(a, b, c, x1, y1); document.write("Image of point (" + x1 + ", " + y1 + ") "); document.write( "by mirror <br> (" + a + ")x + (" + b + ")y + (" + c + ") = 0, is :" ); document.write("(" + x + ", " + y + ")"); </script> |
Image of point (1, 0) by mirror (-1)x + (1)y + (0) = 0, is :(0, 1)
Time Complexity: O(1)
Auxiliary Space: O(1)
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